Linear Programming Applications
Chapter 4 Linear Programming Applications 2.
a.
Let
x1 = units of product 1 produced x2 = units of product 2 produced Max
30x1
+
15x2
x1 0.30x1
+
≤
100
Dept. A
+
0.35x2 0.20x2
≤
36
Dept. B
0.20x1
+
0.50x2
≤
50
Dept. C
s.t.
x1, x2 ≥ 0 Solution: x1 = 77.89, x2 = 63.16 Profit = 3284.21 b.
The dual price for Dept. A is $15.79, for Dept. B it is $47.37, and for Dept. C it is $0.00. Therefore we would attempt to schedule overtime in Departments A and B. Assuming the current labor available is a sunk cost, we should be willing to pay up to $15.79 per hour in Department A and up to $47.37 in Department B.
c.
Let
xA = hours of overtime in Dept. A xB = hours of overtime in Dept. B xC = hours of overtime in Dept. C
Max
30x1
+
15x2
-
18xA
x1 0.30x1
+
0.35x2
-
xA
+
0.20x2
0.20x1
+
0.50x2
-
22.5xB
-
12xC
s.t. -
xB -
xC
xA xB xC x1, x2, xA, xB, xC ≥ 0 x1 = 87.21
≤
100
≤
36
≤
50
≤
10
≤
6
≤
8
Chapter 4
x2 = 65.12 Profit = $3341.34 Overtime Dept. A Dept. B Dept. C
10 hrs. 3.186 hrs 0 hours
Increase in Profit from overtime = $3341.34 - 3284.21 = $57.13
6.
Let
x1 = units of product 1 x2 = units of product 2 b1 = labor-hours Dept. A b2 = labor-hours Dept. B Max
25x1
+ 20x2
+
0b1
6x1 12x1
+
-
1b1
+
0b2
s.t. 8x2 + 10x2
1b1
=
0
-
1b2
=
0
+
1b2
≤
900
x1, x2, b1, b2 ≥ 0 Solution: x1 = 50, x2 = 0, b1 = 300, b2 = 600 Profit: $1,250 8.
Let
x1 = the number of officers scheduled to begin at 8:00 a.m. x2 = the number of officers scheduled to begin at noon x3 = the number of officers scheduled to begin at 4:00 p.m. x4 = the number of officers scheduled to begin at 8:00 p.m. x5 = the number of officers scheduled to begin at midnight x6 = the number of officers scheduled to begin at 4:00 a.m.
The objective function to minimize the number of officers required is as follows: Min
x1 + x2 + x3 + x4 + x5 + x6
The constraints require the total number of officers of duty each of the six four-hour periods to be at least equal to the minimum officer requirements. The constraints for the six four-hour periods are as follows: Time of Day 8:00 a.m. - noon noon to 4:00 p.m.
x1 x1
+ x6 + x2
≥
5
≥
6
Linear Programming Applications
4:00 p.m. - 8:00 p.m.
x2
8:00 p.m. - midnight
+ x3 x3
midnight - 4:00 a.m.
≥ 10 + x4 x4
+ x5
4:00 a.m. - 8:00 a.m.
x5
+ x6
≥
7
≥
4
≥
6
x1, x2, x3, x4, x5, x6 ≥ 0 Schedule 19 officers as follows: x1 = 3 begin at 8:00 a.m. x2 = 3 begin at noon x3 = 7 begin at 4:00 p.m. x4 = 0 begin at 8:00 p.m. x5 = 4 begin at midnight x6 = 2 begin at 4:00 a.m.
11.
Let
xij = units of component i purchased from supplier j
Min
12x11
+
13x1
+ 14x13
+ 10x21
+
11x22
+ 10x23
2 s.t. x11
+
x1
+
x13
= 1000
2 x21 x11
+ x1
+
x22
+
x23
x21 +
=
800
≤
600
≤ 1000
x22
2 x13
+
x23
x11, x12, x13, x21, x22, x23 ≥ 0 Solution: 1 Component 1 Component 2
17. a.
Let
FM FP SM SP TM TP
= = = = = =
600 0
Supplier 2
3
400 0 0 800 Purchase Cost = $20,400
number of frames manufactured number of frames purchased number of supports manufactured number of supports purchased number of straps manufactured number of straps purchased
≤
800
Chapter 4
Min s.t.
38FM 3.5FM 2.2FM 3.1FM FM
+ 51FP
+ 11.5SM + + +
+
+ 15SP
1.3SM 1.7SM 2.6SM
+ 6.5TM
+ 7.5TP
+ 0.8TM + 1.7TM
FP SM
+
SP
TM FM, FP, SM, SP, TM, TP ≥ 0.
+
TP
≤ ≤ ≤ ≥ ≥ ≥
21,000 25,200 40,800 5,000 10,000 5,000
Solution: Frames Supports Straps
Manufacture 5000 2692 0
Purchase 0 7308 5000
b.
Total Cost = $368,076.91
c.
Subtract values of slack variables from minutes available to determine minutes used. Divide by 60 to determine hours of production time used. Constraint 1 2 3
Cutting: Milling: Shaping:
Slack = 0 350 hours used (25200 - 9623) / 60 = 259.62 hours (40800 - 18300) / 60 = 375 hours
d.
Nothing, there are already more hours available than are being used.
e.
Yes. The current purchase price is $51.00 and the reduced cost of 3.577 indicates that for a purchase price below $47.423 the solution may improve. Resolving with the coefficient of FP = 45 shows that 2714 frames should be purchased. The optimal solution is as follows:
OPTIMAL SOLUTION Objective Function Value = Variable -------------FM FP SM SP TM TP
361500.000
Value --------------2285.714 2714.286 10000.000 0.000 0.000 5000.000
Reduced Costs -----------------0.000 0.000 0.000 0.900 0.600 0.000
Linear Programming Applications
Constraint -------------1 2 3 4 5 6
19. a.
Let
Slack/Surplus --------------0.000 3171.429 7714.286 0.000 0.000 0.000
Dual Prices -----------------2.000 0.000 0.000 -45.000 -14.100 -7.500
x11 = amount of men's model in month 1 x21 = amount of women's model in month 1 x12 = amount of men's model in month 2 x22 = amount of women's model in month 2 s11 = inventory of men's model at end of month 1 s21 = inventory of women's model at end of month 1 s12 = inventory of men's model at end of month 2 s22 = inventory of women's model at end of month
The model formulation for part (a) is given. Min
120x11 + 90x21 + 120x12 + 90x22 + 2.4s11 + 1.8s21 + 2.4s12 + 1.8s22
s.t. 20 + x11 - s11 = 150 or x11 - s11 = 130
Satisfy Demand
[1]
Satisfy Demand
[2]
s11 + x12 - s12 = 200
Satisfy Demand
[3]
s21 + x22 - s22 = 150
Satisfy Demand
[4]
30 + x21 - s21 = 125 or x21 - s21 = 95
Labor Hours:
s12
≥ 25
Ending Inventory
[5]
s22
≥ 25
Ending Inventory
[6]
Labor Smoothing for
[7]
Men’s = 2.0 + 1.5 = 3.5 Women’s = 1.6 + 1.0 = 2.6
3.5 x11 + 2.6 x21 ≥ 900 3.5 x11 + 2.6 x21 ≤ 1100
Month 1
[8] 3.5 x11 + 2.6 x21 - 3.5 x12 - 2.6 x22 ≤ 100 [9]
Labor Smoothing for
Chapter 4 -3.5 x11 - 2.6 x21 + 3.5 x12 + 2.6 x22 ≤ 100
Month 2
[10]
x11, x12, x21, x22, s11, s12, s21, s22 ≥ 0 The optimal solution is to produce 193 of the men's model in month 1, 162 of the men's model in month 2, 95 units of the women's model in month 1, and 175 of the women's model in month 2. Total Cost = $67,156 Inventory Schedule 63 Men's 25 Men's
Month 1 Month 2
0 Women's 25 Women's
Labor Levels 1000.00 hours 922.25 hours 1022.25 hours
Previous month Month 1 Month 2 b.
To accommodate this new policy the right-hand sides of constraints [7] to [10] must be changed to 950, 1050, 50, and 50 respectively. The revised optimal solution is given. x11 = 201 x21 = 95 x12 = 154 x22 = 175
Total Cost = $67,175
We produce more men's models in the first month and carry a larger men's model inventory; the added cost however is only $19. This seems to be a small expense to have less drastic labor force fluctuations. The new labor levels are 1000, 950, and 994.5 hours each month. Since the added cost is only $19, management might want to experiment with the labor force smoothing restrictions to enforce even less fluctuations. You may want to experiment yourself to see what happens. 21.
Decision variables : Regular Model Bookshelf Floor
Month 1 B1R F1R
Month 2 B2R F2R
Month 1 B1O F1O
Month 2 B2O F2O
Regular .7 (22) = 15.40
Overtime .7 (33) = 23.10
Decision variables : Overtime Model Bookshelf Floor Labor costs per unit Model Bookshelf
Linear Programming Applications
Floor
1 (22) = 22
1 (33) = 33
IB = Month 1 ending inventory for bookshelf units IF = Month 1 ending inventory for floor model Objective function Min + + + +
15.40 B1R + 15.40 B2R + 22 F1R + 22 F2R 23.10 B1O + 23.10 B2O + 33 F1O + 33 F2O 10 B1R + 10 B2R + 12 F1R + 12 F2R 10 B1O + 10 B2O + 12 F1O + 12 F2O 5 IB + 5 IF
or Min
25.40 B1R + 25.40 B2R + 34 F1R + 34 F2R + 33.10 B1O + 33.10 B2O + 45 F1O + 45 F2O + 5 IB + 5 IF
s.t. .7 B1R + 1 F1R .7 B2R + 1 F2R .7B1O + 1 F1O .7B2O + 1 F2O B1R + B1O - IB IB + B2R + B2O F1R + F1O - IF IF + F2R + F2O
≤ ≤ ≤ ≤ = = = =
2400 2400 1000 1000 2100 1200 1500 2600
Regular time: month 1 Regular time: month 2 Overtime: month 1 Overtime: month 2 Bookshelf: month 1 Bookshelf: month 2 Floor: month 1 Floor: month 2
OPTIMAL SOLUTION Objective Function Value =
241130.000
Variable -------------B1R B2R F1R F2R B1O B2O F1O F2O IB IF
Value --------------2100.000 1200.000 930.000 1560.000 0.000 0.000 610.000 1000.000 0.000 40.000
Reduced Costs -----------------0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.500 0.000
Constraint -------------1 2 3 4
Slack/Surplus --------------0.000 0.000 390.000 0.000
Dual Prices -----------------11.000 16.000 0.000 5.000
Chapter 4
5 6 7 8
0.000 0.000 0.000 0.000
-33.100 -36.600 -45.000 -50.000
OBJECTIVE COEFFICIENT RANGES Variable -----------B1R B2R F1R F2R B1O B2O F1O F2O IB IF
Lower Limit --------------23.900 No Lower Limit 34.000 34.000 33.100 33.100 40.000 No Lower Limit 3.500 0.000
Current Value --------------25.400 25.400 34.000 34.000 33.100 33.100 45.000 45.000 5.000 5.000
Upper Limit --------------25.400 25.400 36.143 50.000 No Upper Limit No Upper Limit 45.000 45.000 No Upper Limit 7.143
Current Value --------------2400.000 2400.000 1000.000 1000.000 2100.000 1200.000 1500.000 2600.000
Upper Limit --------------3010.000 2440.000 No Upper Limit 1040.000 2657.143 1757.143 1890.000 2990.000
RIGHT HAND SIDE RANGES Constraint -----------1 2 3 4 5 6 7 8
23.
Lower Limit --------------2010.000 2010.000 610.000 610.000 1228.571 1142.857 890.000 2560.000
Let
F M A Im Dm sm
= = = = = =
number of windows manufactured in February number of windows manufactured in March number of windows manufactured in April increase in production level necessary during month m decrease in production level necessary during month m ending inventory in month m
Min
1I1 + 1I2 + 1I3 + 0.65D1 + 0.65D2 + 0.65D3
s.t. 9000 + F - s1 = 15,000
February Demand
or (1)
F1 - s1 = 6000
(2)
s1 + M - s2 = 16,500
March Demand
(3)
s2 + A - s3 = 20,000
April Demand
Linear Programming Applications
F - 15,000 = I1 - D1
Change in February Production
or (4)
F - I1 + D1 = 15,000 M - F = I2 - D2
Change in March Production
or (5)
M - F - I2 + D2 = 0 A - M = I3 - D3
Change in April Production
or (6)
A - M - I3 + D3 = 0
(7)
F ≤ 14,000
February Production Capacity
(8)
M ≤ 14,000
March Production Capacity
(9)
A ≤ 18,000
April Production Capacity
(10)
s1 ≤ 6,000
February Storage Capacity
(11)
s2 ≤ 6,000
March Storage Capacity
(12)
s3 ≤ 6,000
April Storage Capacity
Optimal Solution: Cost = $6,450
Production Level Increase in Production Decrease in Production Ending Inventory
February 12,000 0 3,000 6,000
Case Problem 3: Textile Mill Scheduling Let X3R = Yards of fabric 3 on regular looms X4R = Yards of fabric 4 on regular looms X5R = Yards of fabric 5 on regular looms X1D = Yards of fabric 1 on dobbie looms X2D = Yards of fabric 2 on dobbie looms X3D = Yards of fabric 3 on dobbie looms X4D = Yards of fabric 4 on dobbie looms X5D = Yards of fabric 5 on dobbie looms Y1 = Yards of fabric 1 purchased Y2 = Yards of fabric 2 purchased Y3 = Yards of fabric 3 purchased Y4 = Yards of fabric 4 purchased Y5 = Yards of fabric 5 purchased
March 14,000 2,000 0 3,500
April 16,500 2,500 0 0
Chapter 4
Profit Contribution per Yard
Fabric
1 2 3 4 5
Manufactured 0.33 0.31 0.61 0.73 0.20
Purchased 0.19 0.16 0.50 0.54 0.00
1 2 3 4 5
Regular — — 0.1912 0.1912 0.2398
Dobbie 0.21598 0.21598 0.1912 0.1912 0.2398
Production Times in Hours per Yard
Fabric
Model may use a Max Profit or Min Cost objective function. Max
0.61X3R + 0.73X4R + 0.20X5R + 0.33X1D + 0.31X2D + 0.61X3D + 0.73X4D + 0.20X5D + 0.19Y1 + 0.16Y2 + 0.50Y3 + 0.54Y4
or Min
0.49X3R + 0.51X4R + 0.50X5R + 0.66X1D + 0.55X2D + 0.49X3D + 0.51X4D + 0.50X5D + 0.80Y1 + 0.70Y2 + 0.60Y3 + 0.70Y4 + 0.70Y5
Regular Hours Available 30 Looms x 30 days x 24 hours/day = 21600 Dobbie Hours Available 8 Looms x 30 days x 24 hours/day = 5760 Constraints: Regular Looms: 0.192X3R + 0.1912X4R + 0.2398X5R ≤ 21600 Dobbie Looms: 0.21598X1D + 0.21598X2D + 0.1912X3D + 0.1912X4D + 0.2398X5D ≤ 5760 Demand Constraints X1D + Y1 X2D + Y2 X3R + X3D + Y3 X4R + X4D + Y4 X5R + X5D + Y5
= 16500 = 22000 = 62000 = 7500 = 62000
Linear Programming Applications
OPTIMAL SOLUTION Objective Function Value = Variable -------------X3R X4R X5R X1D X2D X3D X4D X5D Y1 Y2 Y3 Y4 Y5 Constraint -------------1 2 3 4 5 6 7
62531.91797
Value --------------27711.29297 7500.00000 62000.00000 4669.13672 22000.00000 0.00000 0.00000 0.00000 11830.86328 0.00000 34288.70703 0.00000 0.00000 Slack/Surplus --------------0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
Reduced Costs -----------------0.00000 0.00000 0.00000 0.00000 0.00000 0.01394 0.01394 0.01748 0.00000 0.01000 0.00000 0.08000 0.06204 Dual Prices -----------------0.57531 0.64821 0.19000 0.17000 0.50000 0.62000 0.06204
OBJECTIVE COEFFICIENT RANGES Variable -----------X3R X4R X5R X1D X2D X3D X4D X5D Y1 Y2 Y3 Y4 Y5
Lower Limit --------------0.50000 0.71606 0.18252 0.31426 0.30000 No Lower Limit No Lower Limit No Lower Limit 0.18000 No Lower Limit 0.48606 No Lower Limit No Lower Limit
Current Value --------------0.61000 0.73000 0.20000 0.33000 0.31000 0.61000 0.73000 0.20000 0.19000 0.16000 0.50000 0.54000 0.00000
Upper Limit --------------0.62394 No Upper Limit No Upper Limit 0.34000 No Upper Limit 0.62394 0.74394 0.21748 0.20574 0.17000 0.61000 0.62000 0.06204
Current Value ---------------
Upper Limit ---------------
RIGHT HAND SIDE RANGES Constraint ------------
Lower Limit ---------------
Chapter 4
1 2 3 4 5 6 7
16301.60059 4751.55957 4669.13672 10169.13672 27711.29297 0.00000 34660.54688
21600.00000 5760.00000 16500.00000 22000.00000 62000.00000 7500.00000 62000.00000
28156.00000 8315.23047 No Upper Limit 26669.13672 No Upper Limit 35211.29297 84095.07813
Production/Purchase Schedule (Yards) Regular Looms
Fabric
1 2 3 4 5
Dobbie Looms 4669 22000
Purchased 11831
27711 7500 62000
34289
Projected Profit: $62,531.92 Value of 9th Dobbie Loom Dual Price (Constraint 2) = 0.64821 per hour dobbie
Monthly Value of 1 Dobbie Loom (30 days)(24 hours/day)($0.64821) = $466.71 Note: This change is within the Right-Hand Side Ranges for Constraint 2. Discussion of Objective Coefficient Ranges For example, fabric one on the dobbie loom shares ranges of 0.31426 to 0.34 for the profit maximization model or 0.64426 to 0.67 for the cost minimization model. Note here that since demand for the fabrics is fixed, both the profit maximization and cost minimization models will provide the same optimal solution. However, the interpretation of the ranges for the objective function coefficients differ for the two models. In the profit maximization case, the coefficients are profit contributions. Thus, the range information indicates how price per unit and cost per unit may vary simultaneously. That is, as long as the net changes in price per unit and cost per unit keep the profit contributions within the ranges, the solution will remain optimal. In the cost minimization model, the coefficients are costs per unit. Thus, the range information indicates that assuming price per unit remains fixed how much the cost per unit may vary and still maintain the same optimal solution.
Case Problem 4: Workforce Scheduling 1.
Let tij = number of temporary employees hired under option i (i = 1, 2, 3) in month j (j = 1 for January, j = 2 for February and so on) The following table depicts the decision variables used in this case problem.
Linear Programming Applications
Option 1 Option 2 Option 3
Jan. t11 t21 t31
Feb. t12 t22 t32
Mar. t13 t23 t33
Apr. t14 t24 t34
May t15 t25
June t16
Costs: Contract cost plus training cost Option 1 2 3
Contract Cost $2000 $4800 $7500
Training Cost $875 $875 $875
Total Cost $2875 $5675 $8375
Min. 2875(t11 + t12 + t13 + t14 + t15 + t16) + 5675(t21 + t22 + t23 + t24 + t25) + 8375(t31 + t32 + t33 + t34) One constraint is required for each of the six months.
Constraint 1: Need 10 additional employees in January t11 = number of temporary employees hired under Option 1 (one-month contract) in January t21 = number of temporary employees hired under Option 2 (two-month contract) in January t31 = number of temporary employees hired under Option 3 (three-month contract) in January t11 + t21 + t31 = 10 Constraint 2: Need 23 additional employees in February t12 , t22 and t32 are the number of temporary employees hired under Options 1, 2 and 3 in February. But, temporary employees hired under Option 2 or Option 3 in January will also be available to satisfy February needs. t21 + t31 + t12 + t22 + t32 = 23 Note: The following table shows the decision variables used in this constraint Jan. Option 1 Option 2 Option 3
t21 t31
Feb. t12 t22 t32
Mar.
Apr.
May
June
Constraint 3: Need 19 additional employees in March
Option 1 Option 2 Option 3
Jan.
Feb.
t31
t22 t32
Mar. t13 t23 t33
Apr.
May
June
Chapter 4
t31 + t22 + t32 + t13 + t23 + t33 = 19 Constraint 4: Need 26 additional employees in May Jan. Option 1 Option 2 Option 3
Feb.
Mar.
t32
t23 t33
Apr. t14 t24 t34
May
June
t32 + t23 + t33 + t14 + t24 + t34 = 26 Constraint 5: Need 20 additional employees in May Jan.
Feb.
Option 1 Option 2 Option 3
Mar.
Apr.
t33
t24 t34
May t15 t25
June
t33 + t24 + t34 + t15 + t25 = 20
Constraint 6: Need 14 additional employees in June Jan.
Feb.
Mar.
Option 1 Option 2 Option 3
Apr.
May
June t16
t25 t34
t34 + t25 + t16 = 14 Optimal Solution: Total Cost = $313,525
Option 1 Option 2 Option 3
Jan. 0 3 7
Feb. 1 0 12
Mar. 0 0 0
Apr. 0 0 14
May 6 0
June 0
2. Option 1 2 3
3.
Number Hired 7 3 33 Total:
Contract Cost $14,000 $14,400 $247,500 $275,900
Training Cost $6,125 $2,625 $28,875 $37,625
Total Cost $20,125 $17,025 $276,375 $313,525
Hiring 10 full-time employees at the beginning of January will reduce the number of temporary employees needed each month by 10. Using the same linear programming model with the right-hand sides of 0, 13, 9, 16, 10 and 4, provides the following schedule for temporary employees:
Option 1 Option 2
Jan. 0 0
Feb. 4 0
Mar. 0 0
Apr. 0 3
May 3 0
June 0
Linear Programming Applications
Option 3
Option 1 2 3 Total:
0
9
Number Hired 7 3 13 23
0
4
Contract Cost $14,000 $14,400 $97,500
Training Cost $6,125 $2,625 $11,375
Total Cost $20,125 $17,025 $108,875 $146,025
Full-time employees cost: Training cost: 10($875) = $8,750 Salary: 10(6)(168)($16.50) = $166,320 Total Cost = $146,025 + $8750 + $166,320 = $321,095 Hiring 10 full-time employees is $321,095 - $313,525 = $7,570 more expensive than using temporary employees. Do not hire the 10 full-time employees. Davis should continue to contract with WorkForce to obtain temporary employees.
4.
With the lower training costs, the costs per employee for each option are as follows: Option 1 2 3
Cost $2000 $4800 $7500
Training Cost $700 $700 $700
Total Cost $2700 $5500 $8200
Resolving the original linear programming model with the above costs indicates that Davis should hire all temporary employees on a one-month contract specifically to meet each month's employee needs. Thus, the monthly temporary hire schedule would be as follows: January - 10; February - 23; March - 19; April - 26; May - 20; and June - 14. The total cost of this strategy is $302,400. Note that if training costs were any lower, this would still be the optimal hiring strategy for Davis.
Case Problem 5: Cinergy Coal Allocation A linear programming model can be used to determine how much coal to buy from each of the mining companies and where to ship it. Let xij = tons of coal purchased from supplier i and used by generating unit j The objective function minimizes the total cost to buy and burn coal. The objective function coefficients, cij , are the cost to buy coal at mine i, ship it to generating unit j, and burn it at generating unit j. Thus, the objective function is ∑ ∑ cij xij . In computing the objective function coefficients three inputs must be added: the cost of the coal, the transportation cost to the generating unit, and the cost of processing the coal at the generating unit. There are two types of constraints: supply constraints and demand constraints. The supply constraints limit the amount of coal that can be bought under the various
Chapter 4 contracts. For the fixed-tonnage contracts, the constraints are equalities. For the variable-tonnage contracts, any amount of coal up to a specified maximum may be purchased. Let Li represent the amount that must be purchased under fixed-tonnage contract i and Si represent the maximum amount that can be purchased under variable-tonnage contract i. Then the supply constraints can be written as follows:
∑x
= Li
for all fixed-tonnage contracts
∑x
≤ Si
for all variable-tonnage contracts
ij
j
ij
j
The demand constraints specify the number of mWh of electricity that must be generated by each generating unit. Let aij = mWh hours of electricity generated by a ton of coal purchased from supplier i and used by generating unit j, and Dj = mWh of electricity demand at generating unit j. The demand constraints can then be written as follows:
∑a
x = Dj
ij ij
i
for all generating units
Note: Because of the large number of calculations that must be made to compute the objective function and constraint coefficients, we developed an Excel spreadsheet model for this problem. Copies of the data and model worksheets are included after the discussion of the solution to parts (a) through (f).
1.
The number of tons of coal that should be purchased from each of the mining companies and where it should be shipped is shown below: Miami Fort Miami Fort #5 #7 RAG
Beckjord
East Bend Zimmer
0
0
61,538
288,462
0
Peabody
217,105
11,278
71,617
0
0
American
0
0
0
0
275,000
Consol
0
0
33,878
0
166,122
Cyprus Addingto n
0
0
0
0
0
0
200,000
0
0
0
Waterloo
0
0
98,673
0
0
The total cost to purchase, deliver, and process the coal is $53,407,243. 2.
The cost of the coal in cents per million BTUs for each generating unit is as follows: Miami Fort #5 111.84
3.
Miami Fort #7 136.97
Beckjor d 127.24
East Bend 103.85
Zimme r 114.51
The average number of BTUs per pound of coal received at each generating unit is shown below:
Linear Programming Applications
Miami Fort #5 13,300
Miami Fort #7 12,069
Beckjor d 12,354
East Bend 13,000
Zimme r 12,468
4.
The sensitivity report shows that the shadow price per ton of coal purchased from American Coal Sales is -$13 per ton and the allowable increase is 88,492 tons. This means that every additional ton of coal that Cinergy can purchase at the current price of $22 per ton will decrease cost by $13. So even paying $30 per ton, Cinergy will decrease cost by $5 per ton. Thus, they should buy the additional 80,000 tons; doing so will save them $5(80,000) = $400,000.
5.
If the energy content of the Cyprus coal turns out to be 13,000 BTUs per ton the procurement plan changes as shown below: Miami Fort Miami Fort #5 #7 Beckjord East Bend RAG
0
0
61,538
288,462
0
Peabody
36,654
191,729
71,617
0
0
American
0
0
0
0
275,000
Consol
0
0
33,878
0
166,122
Cyprus
0
0
85,769
0
0
200,000
0
0
0
0
0
0
0
0
0
Addington Waterloo 6.
Zimmer
The shadow prices for the demand constraints are as follows: Miami Fort #5 21
Miami Fort #7 20
Beckjor d 20
East Bend 18
Zimme r 19
The East Bend unit is the least cost producer at the margin ($18 per mWh), and the allowable increase is 160,000 mWh. Thus, Cinergy should sell the 50,000 mWh over the grid. The additional electricity should be produced at the East Bend generating unit. Cinergy’s profit will be $12 per mWh. The Excel data and model worksheets used to solve the Cinergy coal allocation problem are as follows:
Chapter 4
